An algebra problem before ending 2017

Answer : $\infty$

There is an infinite number of real solutions for this problem.

There are many ways to solve it. Here I write two: Coordinate Geometry and Algebra

Using Coordinate Geometry

We can rearrange the equation into

$(a + \frac{1}{2})^2 + (b + \frac{1}{2})^2 = ( \sqrt{\frac{4035}{2}} )^2$

This is the equation of a circle, centred at $(-\frac{1}{2}, -\frac{1}{2})$ with radius $\sqrt{\frac{4035}{2}}$ . Every point on the circumference of the circle is a solution.

Hence, there should be an infinite number of real solutions. $\square$

Using Algebra

$a^2 +b^2 + a+b= 2017$

$a^2+b^2+a+b-2017=0$

Using quadratic equation,

$a=\frac{-1+\sqrt{8069-4{b}^{2}-4b}}{2},\frac{-1-\sqrt{8069-4{b}^{2}-4b}}{2}$

To make $a$ a real number, the discriminant $8069-4b^2 -4b$ must be $\geq 0$

Using quadratic formula,

$b=\frac{4+4\sqrt{8070}}{-8},\frac{4-4\sqrt{8070}}{-8}$

We then test the intervals.

$b\le -\frac{1+\sqrt{8070}}{2}$

$-\frac{1+\sqrt{8070}}{2}\le b\le -\frac{1-\sqrt{8070}}{2}$

$b\ge -\frac{1-\sqrt{8070}}{2}$

Let us skip the messy algebra of testing the intervals. If you want to look at the testing, click here

Therefore, $-\frac{1+\sqrt{8070}}{2}\le b\le -\frac{1-\sqrt{8070}}{2}$

$b$ has an infinite number of real solutions. Hence, $a$ also has an infinite number of real solutions.

Hence, $a^2 +b^2 + a+b= 2017$ has an infinite number of real solutions. $\square$